Wikipedia:Reference desk/Mathematics: Difference between revisions

 

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:Fun fact: Despite the spelling, “kernel” does not come from German; the closest German equivalent is “der Kern”. German does have “der Kernel” for the core of an operating system, but (ironically) that’s borrowed from English. –[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 01:08, 6 October 2025 (UTC)

:Fun fact: Despite the spelling, “kernel” does not come from German; the closest German equivalent is “der Kern”. German does have “der Kernel” for the core of an operating system, but (ironically) that’s borrowed from English. –[[User:RDBury|RDBury]] ([[User talk:RDBury|talk]]) 01:08, 6 October 2025 (UTC)

:If in a context discussing kernels a restricted variant of function <math>\varphi\!:A\to B</math> is considered, in which the domain is restricted to <math>A’\subsetneq A,</math> the author is well advised not to overload the name <math>\varphi</math> but to denote this variant using <math>\varphi‘{\restriction}_A</math> or to introduce a new name <math>\varphi’\!:A’\to B\,’.</math> When authors follow this advise, <math>\ker_A(\varphi)</math> is the same as <math>\ker_{\text{dom}(\varphi)}\!(\varphi),</math> so then the subscript is redundant. &nbsp;&ZeroWidthSpace;‑‑[[User talk:Lambiam#top|Lambiam]] 11:45, 6 October 2025 (UTC)

:If in a context discussing kernels a restricted variant of function <math>\varphi\!:A\to B</math> is considered, in which the domain is restricted to <math>A’\subsetneq A,</math> the author is well advised not to overload the name <math>\varphi</math> but to denote this variant using <math>\varphi{\restriction}_A</math> or to introduce a new name <math>\varphi’\!:A’\to B\,’.</math> When authors follow this advise, <math>\ker_A(\varphi)</math> is the same as <math>\ker_{\text{dom}(\varphi)}\!(\varphi),</math> so then the subscript is redundant. &nbsp;&ZeroWidthSpace;‑‑[[User talk:Lambiam#top|Lambiam]] 11:45, 6 October 2025 (UTC)

= October 8 =

= October 8 =

-1/12 is the sum of all positive integers, but what is the product of all positive integers? 61.229.98.9 (talk) 21:42, 28 September 2025 (UTC)[reply]

First of all, no, it’s a divergent sum with no finite value. There are certain very generalized meanings of “sum” where the value turns out to be -1/12 (see Riemann zeta function#Specific values), but it’s not the sum in the generally accepted sense. Similarly, the product diverges and has no finite value. There may be a way to do a similar generalized version of a product, but AFIAK there isn’t a lot of interest in that kind of thing and there’s no reason to expect there would be a meaningful value at the end of the process. —RDBury (talk) 00:30, 29 September 2025 (UTC)[reply]

The standard regularization of the product of the positive integers, analogous to the regularization of ζ ( − 1 ) = − 1 / 12 {\displaystyle \zeta (-1)=-1/12} is e − ζ ′ ( 0 ) = 2 π {\displaystyle e^{-\zeta ‘(0)}={\sqrt {2\pi }}} . I edited an article describing this, but can’t find it now. You’re welcome to search my contributions for it. (Of course the naive regularization, via the gamma function, has an essential singularity at infinity. ) Tito Omburo (talk) 00:54, 29 September 2025 (UTC)[reply]

Mathematically speaking, does a Mercator projection have to be oriented so that the actual poles are at the top and bottom? Obviously it wouldn’t be very useful, but would it be possible to produce a Mercator projection with random spots as poles and a random line on the surface (geometrically, a plane that’s perpendicular to the line connecting the polar points) as an equator? I assume so, but I could be wrong, and I don’t know what to call it; Mercator projection doesn’t address “alternate” projections. Nyttend (talk) 03:42, 29 September 2025 (UTC)[reply]

The article has sections on Transverse Mercator and Oblique Mercator, each with its own main article. Isn’t that what you’re thinking of? —Wrongfilter (talk) 05:20, 29 September 2025 (UTC)[reply]

Oops, yes it is. I skipped the “uses” section because it covers things like marine navigation; I didn’t anticipate that types of the projection would be thrown in there too. Nyttend (talk) 09:20, 29 September 2025 (UTC)[reply]

For completeness, Transverse and Oblique Mercator require the poles to be antipodes, not just any two “random spots”. At the extreme, if the spots coincide, one gets an azimuthal projection. cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 08:32, 30 September 2025 (UTC)[reply]

A Möbius transformation exists to map any pair of points to the poles! —Tamfang (talk) 21:28, 6 October 2025 (UTC)[reply]

On my last air trip, there was a seat-back display of our progress, on a standard Mercator map; I wished it was oblique instead! —Tamfang (talk) 21:30, 6 October 2025 (UTC)[reply]

Publishers typically sort their Sudoku puzzles into difficulties: easy, medium, hard, expert – or something like that. Is that process manual/subjective – or is there an algorithm that’s able to sort the required moves into different types? For example, you might say that easy puzzles only require straightforward comparison checks, while medium puzzles might require some inferential checks or row/column based checks, while harder puzzles would require more complicated inferences (I assume there’s proper jargon for this stuff, but hopefully I’m getting the gist across). The actual usage is of interest, but I’m more curious about the theoretical side, which is why I’m asking here rather than the Ent desk. Matt Deres (talk) 15:07, 30 September 2025 (UTC)[reply]

A couple of years ago I looked for material on this but came out empty-handed.

I thought of using an algorithm that imitates a plausible way in which humans solve these puzzles and counts the number of inferential steps but have not carried this through.

For one published series of puzzles, with difficulties ranging from 1 to 4 stars, I can report (from personal experience) that levels 1 and 2 are boringly easy, while there is substantial increase in difficulty from 2 to 3 (measured by the time needed to solve them, which goes up by a factor of 3, or so). But occasionally a 3-star puzzle is easier than the typical 2, or a 2-star puzzle proves more challenging than the typical 3. There is also a jump from 3 to 4, but not by that large a factor as from 2 to 3. Furthermore, there is no common difficulty measure like the Scoville scale for hotness; one publisher’s “medium” may be easier than another publisher’s “easy”.  ​‑‑Lambiam 17:05, 30 September 2025 (UTC)[reply]

Yeah, that’s broadly my experience as well and it was the “expert” ones that were perhaps only “hard” that prompted my question. In practice, I assume such stuff often comes down to happening to notice a particular item ahead of when it would more typically be found during a brute-force and it gives you an outsize advantage to complete the puzzle. Matt Deres (talk) 20:04, 30 September 2025 (UTC)[reply]

Spotting something ahead of when a systematic approach would have found it can help, but actively looking for such things may not pay off and slow one down. Also, in general you have to alternate between two modes, which I refer to as “digit seeks cell” and “cell seeks digit”. I see no clear criterion when to choose which, although “cell seeks digit” is generally less fruitful in sparse puzzles. I have a rule of thumb to avoid unsuccessful “cell seeks digit” efforts, which often works but sometimes fails miserably in either direction. The choice is mostly a gamble, though, in which one can be lucky or unlucky.  ​‑‑Lambiam 23:43, 30 September 2025 (UTC)[reply]

Agreed. I’m not suggesting that going snipe hunting is the right thing to do; I’m just saying that sometimes you can stumble across a snipe and find that the expert puzzle is surprisingly easy. I think I understand what you mean by “digit seeks cell” and so on, though it raises the related question of whether the various processes players use have semi-formalized names. There are at least three very different methods I end up using for extreme puzzles (more than that, if we separate out rotations as discrete phases). If that terminology exists, it would be easier to search for whether puzzle-makers take them into account when setting their difficulty levels. Matt Deres (talk) 18:38, 1 October 2025 (UTC)[reply]

If you look at Simon Tatham’s implementations of various standard puzzles [1] (sudoku is there under the name “Solo”) I think what you find is that the program generates puzzles that can definitely be solved with certain collections of rules, or with certain search-depths, and the difficulty level is related to how many rules / how deep a search to allow. —JBL (talk) 23:52, 2 October 2025 (UTC)[reply]

Wikipedia’s article Tetration says there is only one kind of tetration. But the talk page is talking about “lower tetration”, which implies that there is more than one kind. Is this really true?? Georgia guy (talk) 16:05, 2 October 2025 (UTC)[reply]

I think it is a term invented by the poster. Note that, using left-associative grouping, we have

( ⋯ ( ( a 0 a 1 ) a 2 ) ⋅ ⋅ ⋅ ) a n = a 0 a 1 a 2 ⋯ a n , {\displaystyle \left(\cdots {{{\left(\left({a_{0}}^{a_{1}}\right)^{a_{2}}\right)^{\cdot }}^{\cdot }}^{\cdot }}\right)^{a_{n}}=a_{0}^{a_{1}a_{2}\cdots a_{n}},}

so this not very interesting.  ​‑‑Lambiam 17:59, 2 October 2025 (UTC)[reply]

A unitary matrix can be transformed into a unistochastic matrix by applying the norm-squared to each of its elements. Can a similar transformation be applied to more general unitary operators? 203.17.151.90 (talk) 11:26, 5 October 2025 (UTC)[reply]

1. In group theory, we notate the neutral element by e {\displaystyle {\text{e}}} , from the german word einheit.
In ring theory, we notate the additional and multiplicational neutral elements by 0 R , 1 R {\displaystyle 0_{_{R}},1_{_{R}}} respectively (in a unit ring).
Why not notate them e R , u R {\displaystyle {\text{e}}_{_{R}},{\text{u}}_{_{R}}} respectively, with the second from the english word unit?
2. Also, given a general transformation φ : A → B {\displaystyle \varphi :A\to B} , in my humble opinion the notations ker ⁡ ( φ ) , Im ( φ ) {\displaystyle \ker(\varphi ),{\text{Im}}(\varphi )} for both the kernel and image respectively lack a clear reference to the domain and range.
Why not notate them as follows?

ker A ( φ ) {\displaystyle \ker _{_{A}}\!(\varphi )} (read: the kernel within A {\displaystyle A} of the transformation φ {\displaystyle \varphi } )

Im B ( φ ) {\displaystyle {\text{Im}}_{_{B}}\!(\varphi )} (read: the image within B {\displaystyle B} of the transformation φ {\displaystyle \varphi } )

יהודה שמחה ולדמן (talk) 22:49, 5 October 2025 (UTC)[reply]

Notation is not standardized in math, in other words you can use whatever notation you want as long as readers know what you’re talking about. So the notation e for the identity in a group is just a norm, not a rule. Math also allows context to be used to interpret its notation. This introduces a certain amount of ambiguity, but if the meaning is clear from context then there is no harm done, and dropping unneeded subscripts and whatnot actually makes the notation easier to read. In group theory, the identity in the subgroup of a group must be the same as the identity of the group, a fact which must be proved, though that is trivial, so in almost all cases there’s no need to use subscripts to distinguish them. The group operation is specific to the group, not the set of it’s elements. So the unambiguous statement of the associative law is (a ⋅_G b) ⋅_G c = a ⋅_G (b ⋅_G c), which is less clear and a lot more work to write than (ab)c = a(bc). Similarly, the subscripts on Ker and Im might be added to make the notation unambiguous, but if the meaning is clear from context there is no need to, and in most cases it only makes the notation harder to read. You can use subscripts if needed, but I’m pretty sure that situation doesn’t come up much.

Fun fact: Despite the spelling, “kernel” does not come from German; the closest German equivalent is “der Kern”. German does have “der Kernel” for the core of an operating system, but (ironically) that’s borrowed from English. —RDBury (talk) 01:08, 6 October 2025 (UTC)[reply]

If in a context discussing kernels a restricted variant of function φ : A → B {\displaystyle \varphi \!:A\to B} is considered, in which the domain is restricted to A ′ ⊊ A , {\displaystyle A’\subsetneq A,} the author is well advised not to overload the name φ {\displaystyle \varphi } but to denote this variant using φ ↾ A {\displaystyle \varphi {\upharpoonright }_{A}} or to introduce a new name φ ′ : A ′ → B ′ . {\displaystyle \varphi ‘\!:A’\to B\,’.} When authors follow this advise, ker A ⁡ ( φ ) {\displaystyle \ker _{A}(\varphi )} is the same as ker dom ( φ ) ( φ ) , {\displaystyle \ker _{{\text{dom}}(\varphi )}\!(\varphi ),} so then the subscript is redundant.  ​‑‑Lambiam 11:45, 6 October 2025 (UTC)[reply]

In this page look at the image at Probability#Mutually_exclusive_events The image with top and bottom double-circles. So, the mistake is the top 2 circles are called “independent events.” That is a mistake. Can anyone tell why? 24.136.10.82 (talk) 16:04, 8 October 2025 (UTC).[reply]

A and B are independent when P(A∩B) = P(A)P(B). If P(A) and P(B) are both > 0 then the product is > 0, and P(A∩B) has non-zero probability. But if A∩B = ∅ then it’s probability is 0, so A∩B can’t be empty. The top heading should say “Mutually Exclusive Events”. Another issue with the image is that it’s a JPG when SVG is more appropriate for this kinds of diagram. Ideally someone will create an SVG with a corrected title, but in the mean time my vote is to remove the image from the article. I’m not convinced it’s all that helpful anyway. —RDBury (talk) 16:50, 8 October 2025 (UTC)[reply]

Yea so basically there’s a 2 by 2 variables alone. Mutually-exclusive vs. non-mutually exclusive. As well as independent vs. dependent. So you got

not mutually exclusive and dependent: flipping a coin multiple times. This is Venn diagram with 2 overlapping circles.

mutually exclusive and independent: you have a bag of apple, strawberry, and onion. It is possible to draw something that is a fruit and berry at the same time, but not possible to draw something that is a berry and vegetable at same time. This Venn diagram is with 2 circles with no overlap.

not mutually exclusive and dependent: known amount of red and blue marbles in a bag, and pulling marbles out consecutively. This has the same Venn Diagram of 2 circles with overlap, however, the overlap P(A intersect B) is not P(A)*P(B), the intersect is either larger or smaller. Unlike the other, P(A)P(B) = P(A intersect B).

mutually exclusive and independent: not possible, because 1 event is impossible. I.e., rolling a dice twice, and getting a 7 on a try. The Venn diagram is in theory 2 circles non-overlap, but actually looks like 1 circle, because 1 circle has an area of 0, effectively making no overlap. 24.136.10.82 (talk) 18:16, 8 October 2025 (UTC).[reply]

The old diagram is totally confused. P ( A ) {\displaystyle P(A)} is a real number in the unit interval and not a subset of Ω . {\displaystyle \Omega .} I have replaced it with a hopefully less misleading version.  ​‑‑Lambiam 20:34, 8 October 2025 (UTC)[reply]